Mathematics, 09.01.2020 11:31 jordan84546
Use implicit differentiation to find an equation of the tangent line to the curve at the indicated point.
y = sin 3xy; (π/6,1)
Answers: 3
![\bf y=sin(3xy)\qquad \left(\frac{\pi }{6},1 \right)\\\\ -------------------------------\\\\ \cfrac{dy}{dx}=cos(3xy)\left[3\left[ y+x\frac{dy}{dx} \right] \right]\implies \cfrac{dy}{dx}=cos(3xy)\left[3y+3x\frac{dy}{dx} \right] \\\\\\ \cfrac{dy}{dx}=3ycos(3xy)+3xcos(3xy)\cfrac{dy}{dx} \\\\\\ \cfrac{dy}{dx}-3xcos(3xy)\cfrac{dy}{dx}=3ycos(3xy)\impliedby \textit{common factor}](/tpl/images/0448/4241/eec61.png)
![\bf \cfrac{dy}{dx}[1-3xcos(3xy)]=3ycos(3xy)\implies \boxed{\cfrac{dy}{dx}=\cfrac{3ycos(3xy)}{1-3xcos(3xy)}} \\\\\\ \left. \cfrac{3ycos(3xy)}{1-3xcos(3xy)} \right|_{\frac{\pi }{6},1}\implies \cfrac{3\cdot 1\cdot cos\left( 3\cdot \frac{\pi }{6}\cdot 1 \right)}{1-3\cdot \frac{\pi }{6}cos\left(3\cdot \frac{\pi }{6}\cdot 1 \right)}\\\\\\ \cfrac{3\cdot 1\cdot cos\left( \frac{\pi }{2} \right)}{1-\frac{\pi }{2}cos\left( \frac{\pi }{2} \right)}=0\\\\ -------------------------------\\\\](/tpl/images/0448/4241/f78ec.png)
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